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+--- Thread: Improper integrals at MSE (/showthread.php?tid=943)
Improper integrals at MSE - tommy1729 - 12/21/2014
Last Friday while playing chess I talked to my MSE friend mick.
You might like this :
http://math.stackexchange.com/questions/1076074/about-fractional-iterations-and-improper-integrals
regards
tommy1729
RE: Improper integrals at MSE - tommy1729 - 12/23/2014
Sheldon's answer on MSE is nice.
Thank you Sheldon.
I made an intresting observation relating things to complex dynamics.
The main thing is the mysterious looking change \( x \) -> \( x - 2^t / x. \)
My observation can be considered positive or negative , intresting or dissapointing , it depends on taste I guess and the hope for nontrivial analogues.
But the idea of having some function \( g(x) \) for which every real iterate \( g^{[t]}(x) \) " works " is found , though it might not be as nontrivial as mick hoped. ( not saying a nontrivial case cannot exist ).
- Maybe variants of this exist in calculus textbooks / papers but its very " dynamical " in nature -
Anyway here it is :
\( f_x(t) = x - 2^t / x \)
\( f_x(f_x^{[-1]}(t) + 1) \)
=> \( x - 2^T / x \) with \( T = f_x^{[-1]}(t) + 1 \)
=> \( x - 2/x * 2^{f_x^{-1} (t)} \)
=> \( x - 2/x * Solve(q,x - q/x = t) \)
Solve .. => \( q = x(x-t) \)
Thus :
\( x - 2/x *x(x-t) = x - 2(x-t) = x - 2x + t = -x + t \)
Which is trivial.
Reminds me of this quote :
" Young man, in mathematics you don't understand things. You just get used to them. "
John von Neumann.
Btw I considered doing the things (steps above) in reverse : showing \( x \) -> \( x - t/x \) is valid from the validity of \( x -> -x + t. \)
regards
tommy1729
" the master "
RE: Improper integrals at MSE - sheldonison - 12/24/2014
(12/23/2014, 11:31 PM)tommy1729 Wrote: ....function \( g(x) \) for which every real iterate \( g^{[t]}(x) \) " works " is found , though it might not be as nontrivial as mick hoped. ( not saying a nontrivial case cannot exist ).
...
\( f_x(t) = x - 2^t / x \)
\( f_x(f_x^{[-1]}(t) + 1) \)
=> \( x - 2^T / x \) with \( T = f_x^{[-1]}(t) + 1 \)
=> \( x - 2/x * 2^{f_x^{-1} (t)} \)
=> \( x - 2/x * Solve(q,x - q/x = t) \)
Solve .. => \( q = x(x-t) \)
Thus :
\( x - 2/x *x(x-t) = x - 2(x-t) = x - 2x + t = -x + t \)
....
Hey Tommy,
Not sure I understood all of that ... But it inspired me to consider the following sequence of functions
\( f(x)=x-\frac{1}{2x}\;\;\; g(x)=f^{o2}(x) \)
\( f(x)=x-\frac{1}{4x}\;\;\; g(x)=f^{o4}(x)\;\;\; \)
\( f(x)=x-\frac{1}{8x}\;\;\; g(x)=f^{o8}(x)\;\;\; \)
\( f(x)=x-\frac{1}{16x}\;\;\; g(x)=f^{o16}(x)\;\;\; \)
...
\( \lim_{n \to \infty} f(x)=x-\frac{1}{2^nx}\;\;\; g(x)=f^{o2^n}(x) \)
Does g(x) converge, and is it a solution of interest to Mick? If g(x) converges, and it is analytic, then it has a Taylor/Laurent series....
Update:, by brute force, using a lot of computer cycles to estimate the limit, and then turn the coefficents it back into a fraction with power's of 2's... I get the following Laurent series, as the function that Mick might be looking for.
\( g(x)= x -
\frac{1}{x} -
\frac{1}{2x^{3}} -
\frac{1}{2x^{5}} -
\frac{5}{8x^{7}} -
\frac{7}{8x^{9}} -
\frac{21}{16x^{11}} -
\frac{33}{16x^{13}} -
\frac{429}{128x^{15}} -
\frac{715}{128x^{17}} -
\frac{2431}{256x^{19}} -
\frac{4199}{256x^{21}} -
\frac{29393}{1024x^{23}} -
\frac{52003}{1024x^{25}} -
\frac{185725}{2048x^{27}} -
\frac{334305}{2048x^{29}} - ... \)
It would probably be normally expressed as \( f(x)=\frac{1}{g(1/x) \)
\( f(x) = x + x^3 +
\frac{3 x^5 }{2 } +
\frac{5 x^7 }{2 } +
\frac{35 x^9 }{8 } +
\frac{63 x^{11} }{8 } +
\frac{231 x^{13} }{16 } +
\frac{429 x^{15} }{16 } +
\frac{6435 x^{17} }{128 } +
\frac{12155 x^{19} }{128 } +
\frac{46189 x^{21} }{256 } +
\frac{88179 x^{23} }{256 } +
\frac{676039 x^{25} }{1024 } +
\frac{1300075 x^{27} }{1024 } + ... \)
update2:
This would be compactly expressed via the Abel function as:
\( \alpha(z)=\frac{-1}{2z^2}\;\;\;\alpha(f(z))=\alpha(z)+1\;\;\;\alpha^{-1}(z)=-\sqrt{\frac{-1}{2z}} \)
And then we get:
\( f(z)=\alpha^{-1}(\alpha(z)+1) \;=\; \frac{z}{\sqrt{1-2z^2} \)
Finally, Mick's desired function in closed form would be as follows. With a little algebra, we generate all of the fractional iterates of g(z) as well. Then, using Mick's notation we have the desired g(z,t) function, which has all fractional iterates defined as:
\( g(z,t)=\; \frac{1}{f(1/z)}\; = \; sqrt{z^2-2t}\; = \; z -
\frac{t}{z} -
\frac{t^2}{2z^{3}} -
\frac{t^3}{2z^{5}} -
\frac{5t^4}{8z^{7}} -
\frac{7t^5}{8z^{9}} -
\frac{21t^6}{16z^{11}} - ... \;\;\; \) for t=1, this is the same as the Laurent series above
RE: Improper integrals at MSE - tommy1729 - 12/28/2014
I once had a few threads here were I discussed the need for limits of the form ( a + f(n)/(bn)) ^[n] = C or similar.
This seems very much like your limit, maybe you got inspired from me too.
Anyways I must say that thread was looking for tetration type functions / limits so in that sense your limit is more " classical ".
I was not able to find the threads again but this one is somewhat similar :
http://math.eretrandre.org/tetrationforum/showthread.php?tid=262&highlight=limit
regards
tommy1729
RE: Improper integrals at MSE - tommy1729 - 01/11/2015
This leads to the question
Is there a solution F(x) = super of ( x + a - 2^t/(x+b) ) for some real a,b (super with respect to t) such that integral f(F(x)) dx = integral f(x) ?
regards
tommy1729
RE: Improper integrals at MSE - tommy1729 - 01/16/2015
I think this is an underrated thread.
If it turns out true , this gives an intresting connection between dynamical systems and calculus !
regards
tommy1729